Respuesta :
The pattern is growing horizontally and vertically, with the shape of a ladder. Each square is a step of the ladder. Is symmetrical in its diagonal.
For figure 1, we have 1 square.
For figure 2, we have 2+1 = 3 squares.
For figure 3, we have 3+2+1 = 6 squares.
For figure 4, we have 4+3+2+1 = 10 squares.
We can also look at the pattern as joining a "tower" of squares of size "n" for figure "n".
Figure n will have:
[tex]numberofsquares_n=\sum ^n_{i=1^{}}i=\frac{n(n+1)}{2}[/tex]For figure 10 we have: 10*11/2 = 110/2 = 55 squares.
It will look like a triangle with 10 squares in the base and 10 squares in the left side.
For figure 55 we will have 1540 squares.
[tex]\begin{gathered} n=55 \\ \frac{n(n+1)}{2}=\frac{55\cdot56}{2}=\frac{3080}{2}=1540 \end{gathered}[/tex]To know if 190 correspond to one figure we can calculate:
[tex]\begin{gathered} \frac{n(n+1)}{2}=190 \\ n(n+1)=190\cdot2 \\ n^2+n=380 \\ n^2+n-380=0 \\ \\ n=\frac{-1\pm\sqrt{1-4\cdot1\cdot(-380)}}{2\cdot1}=\frac{-1}{2}\pm\frac{\sqrt{1+1520}}{2} \\ \\ n=\frac{-1}{2}\pm\frac{\sqrt{1521}{}}{2}=\frac{-1}{2}\pm\frac{39}{2} \\ \\ n_1=\frac{-1+39}{2}=\frac{38}{2}=19 \\ n_2=\frac{-1-39}{2}=\frac{-40}{2}=-20 \\ \end{gathered}[/tex]We have a solution for n=19, so in figure 19 we will have a ladder with 190 squares, as it satisfy the pattern we have figured.
